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Manipal Entrance Test 2018 Math Paper

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Question : 44 of 70

Marks: +1, -0

The function

\sin x(1+\cos x), 0 \le x \le \frac{\pi}{2}

has maximum value, when

x

is equal to

0
$\frac{\pi}{2}$
$\frac{\pi}{6}$
None of these

Solution:

Let

z=\sin x(1+\cos x)

or

z=\sin x+\frac{1}{2}\sin 2x

On differentiating w.r.t. x, we get

\frac{dz}{dx}=\cos x+\cos 2x

For maximum or minimum, put

\frac{dz}{dx}=0

\Rightarrow \cos x+\cos 2x=0

\Rightarrow 2\cos^2 x-1+\cos x=0

\Rightarrow (\cos x+1)(2\cos x-1)=0

\Rightarrow \cos x=-1, \frac{1}{2}

\Rightarrow x=\frac{\pi}{3}, \pi

But

0 \le x \le \frac{\pi}{2},

therefore we take only

x=\frac{\pi}{3}

\therefore \frac{d^2 z}{dx^2}=-\sin x-2\sin 2x=-

ve, for

x=\frac{\pi}{3}

∴ It is maximum at

x=\frac{\pi}{3}

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